A) \[C{{H}_{4}}\]
B) \[S{{F}_{4}}\]
C) \[BF_{4}^{-}\]
D) \[NH_{4}^{+}\]
Correct Answer: B
Solution :
When the number of hybrid orbitals, H is 4, the hybridization is sp3. |
\[H=\frac{1}{2}[V+M-C+A]\] where, V = number of valence electrons of central atom |
M = number of monovalent atoms |
C = total positive charge |
A = negative charge |
[a]\[C{{H}_{4}},\] |
\[H=\frac{1}{2}[4+4-0+0]=4,\]thus \[s{{p}^{3}}\]hybridization |
[b] For\[S{{F}_{4}},\] |
\[H=\frac{1}{2}[6+4-0+0]=5,\]thus |
Hybridization |
[c] For \[BF_{4}^{-},\] |
\[H=\frac{1}{2}[3+4-0+1]=4,\]thus\[s{{p}^{3}}\] |
Hybridization |
[d] For\[NH_{4}^{+},\] |
\[H=\frac{1}{2}[5+4-1+0]=4,\]thus\[s{{p}^{3}}\] |
Hybridization |
Thus, only in \[S{{F}_{4}},\] the central atom does not have sp3 hybridization. |
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